Friday, March 21, 2014

Geometry Problem 995: Quadrilateral, Perpendicular Diagonals, Perpendicular Bisector, Parallel

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 995.

Online Geometry Problem 995: Quadrilateral, Perpendicular Diagonals, Perpendicular Bisector, Parallel.

2 comments:

  1. Since ∆AEC and ∆BFC are isosceles triangles
    So ∠BFD=180- 2. ∠ODA
    And ∠CEA=180-2. ∠CAE
    And ∠BFD+∠CEA= 360-2.( ∠ODA+∠CAE)= 180
    So BF//CE

    ReplyDelete
  2. Problem 995
    Let <EAC=<ECA=x,<FBD=<FDBX+Y=y.But in triangle AOD apply <DAO+<ADO=90 or x+y=90. If BF intersects the AC at K then <BKO=90-<KBO=90-y=x=<AKF=<ACF. So
    BF//CE.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete